84
3 Analysis of Strain
J 2 = ε x ε y + ε y ε z + ε z ε x −
1
4
γ
2
xy + γ
2
yz + γ
2
zx
= (0.1) (−0.05) + (−0.05) (0.05) + (0.05) (0.1)
−
1
4
(0.3)
2
+ (0.1)
2
+ (−0.08)
2
∴ J 2 = −0.0291
J 3 = (0.1)(−0.05)(0.05) +
1
4
[(0.3)(0.1)(−0.08) − 0.1(0.1)
2
+ 0.05(0.08)
2
− 0.05(0.3)
2
J 3 = −0.002145
∴ The cubic equation is
ε
3
− 0.1ε
2
− 0.0291ε + 0.002145 = 0
( i )
Now cos 3θ = 4 cos
3
θ − 3 cos θ
Or cos
3
θ −
3
4
cos θ −
1
4
cos 3θ = 0
( i i )
Let
ε = r cos θ +
J 1
3
= r cos θ +
0.1
3
ε = r cos θ + 0.033
.
∴ (i) can be written as
(r cos θ + 0.033)
3
− 0.1(r cos θ + 0.033)
2
− 0.0291(r cos θ + 0.033) + 0.002145
= 0
(r cos θ + 0.033) (r cos θ + 0.033)
2
− 0.1(r cos θ + 0.033)
2
− 0.0291r cos θ
−0.00096 + 0.002145 = 0
(r cos θ + 0.033)
r
2 cos
2
θ + 0.067r cos θ + 0.00109
−0.1
r
2 cos
2
θ + 0.067r cos θ + 0.00109
− 0.0291r cos θ − 0.00096 + 0.002145
= 0
r
3 cos
3
θ + 0.067r
2 cos
2
θ + 0.00109r cos θ + 0.033r
2 cos
2
θ + 0.0022r cos θ
+0.000036 − 0.1r
2 cos
2
θ − 0.0067r cos θ − 0.000109 − 0.0291r cos θ − 0.00096
+0.002145 = 0
i.e., r
3 cos
3
θ − 0.03251r cos θ − 0.00112 = 0
or cos
3
θ −
0.03251
r 2
cos θ −
0.00112
r 3
= 0
(iii)